one no is 7 more than another and its square is 77 more than the square of the smaller no. what are the nos
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Answered by
2
the numbers are x and y .
it is given that x= 7+y ...(1)
it is also given that x² = 77+y²....(2)
from (1) and (2),
(7+y)²=77+y²
we can use the identity (a+b)²=a² +2ab+b²
49+14 y+y² = 77+ y²
y² term gets cancelled
14 y=77 - 49
14 y = 28
y = 28÷14
∴y = 2
also x = 7+ y
x = 9
hence the numbers are 2 and 9
it is given that x= 7+y ...(1)
it is also given that x² = 77+y²....(2)
from (1) and (2),
(7+y)²=77+y²
we can use the identity (a+b)²=a² +2ab+b²
49+14 y+y² = 77+ y²
y² term gets cancelled
14 y=77 - 49
14 y = 28
y = 28÷14
∴y = 2
also x = 7+ y
x = 9
hence the numbers are 2 and 9
khushi2563:
can u solve it using x only??
why?
no it isn't possible as x and y have different values and also x is not equal to y
kk....ty
what is kk.. ty?
Answered by
2
Step-by-step explanation:
Let one number be x
other no =x+7
(x+7)² -77 =x²
x²+49+14x-77=x²
x²-28+14x=x²
x²-x²+14x = 28
14x= 28
x= 2
other no = x+7 =2+7 =9
the numbers are 2 and 9
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